i learned 4 weeks of my senior college physics from this man in about 2 hours. Why can't my professors be this good? Why am I paying thousands of dollars a quarter for an education I just learn on youtube anyways...
Just think of it this way your not necessarily paying for the education what you are paying for is the chance to earn that fancy piece of paper that serves as proof that you know what your doing.
I totally agree with you, bro. My professor just talks following the PPT without any useful information. She only knows to give us a lot of hard tutorial sheets...
In the example u gave at 3:45.If boty v+ and v- are 1.0v.Shouldnt Vout=0 Since Vout=Vin*A And Vin= V+ -(V-) If both are 1 then Vin=0 and so should Vout
HollowHandPuppet it would amplify the difference if there was no feedback. If there’s feedback then output and inverting input monitor and adjust each other constantly
That must be a mistake he made right? Im pretty sure that a non-inverting opamp set up this way where R1=R2 would generate some kind of sinus-wave(ac) at the v0 node because of this "paradox". v0 would go from 2v to 0v to 2v and then back again, oscillating?
@@eyeballs97 assuming v+ was 1v and v- was 1v-6uV (there is a very very small difference between v+ and v-), then the amplified output would be 6v (assuming A=1,000,000). In the example, if Vo increased, it caused v- to increase, which will then cause Vo to decrease back to 6V (Vo's original value), this then makes v- decrease back to 1-6uV. At this point all voltages are back to its initial condition, so there wont be any oscillations. Only when Vo is once affected, then will it cause v- to change accordingly.
So, since the difference on the inputs is amplified and by a huge number, do you need R1=R2 almost exactly for feedback to ensure that V+ = V- and you don't this cause saturation?
i don't get it. Why would u_o go up when the input voltage difference is 0. since the input and output voltage are linear with each other, u_o should become 0 and then you would have this oscillation of values back and forth. please explain
So what happens after Vout decreases would it not just cause V- to decrease causing Vout to increase and then just be stuck in a loop of increasing and decreasing?
You have the idea right. Any change to Vout causes V- to change, which pushes Vout in the opposite direction of its change. That is the essence of feedback. In real life, Vout will settle at exactly the voltage you want and will stay there. (It may waver a tiny bit, but too small to notice.)
v+ and v- are APPROXIMATELY 0 due to the virtual ground theory but they are actually a very very small number... but when the difference between the small number is multiplied by A then we get Vout.
ATM Jahid Hasan it's not exactly, remember that the Vo=2Vm was obtained by approximating A as large (see the previous video) so that we ignored a small corrective term the actual v- differes from 1 by something of order 1/A which is very small.
i learned 4 weeks of my senior college physics from this man in about 2 hours. Why can't my professors be this good? Why am I paying thousands of dollars a quarter for an education I just learn on youtube anyways...
Just think of it this way your not necessarily paying for the education what you are paying for is the chance to earn that fancy piece of paper that serves as proof that you know what your doing.
you're paying to take the tests.
And also paying to see those chicks around your campus...
I totally agree with you, bro. My professor just talks following the PPT without any useful information. She only knows to give us a lot of hard tutorial sheets...
I learn only topics of the lessons from my lecturers.
In the example u gave at 3:45.If boty v+ and v- are 1.0v.Shouldnt Vout=0
Since Vout=Vin*A
And Vin= V+ -(V-)
If both are 1 then Vin=0 and so should Vout
I understood more from your 10 minutes video than many hours of lectures in my electronics class .thank you so much😊🙌🙌🙌🙌
gootno
the video was 6 mins
Thirukumara racimo
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Thirukumara
Thank you for clearly explaining this very critical concept of negative feedback within 6 minutes.
it is still not an easy stuff to understand . Thank you
It’s not all that bad. If you’re still having troubles RSD Academy has a simple and straight forward 9 part opamp playlist.
I finally understand why we are doing all this analysis, thank you!
I kind of knew how it works but somehow not really... Your explanation is excellent! Thank you.
but if the op-amp amplifies the difference between v+ and v-, and in this case case we have v+=1v and v-=1v, shouldn't be the Vout equal to 0 v?
HollowHandPuppet it would amplify the difference if there was no feedback. If there’s feedback then output and inverting input monitor and adjust each other constantly
That must be a mistake he made right? Im pretty sure that a non-inverting opamp set up this way where R1=R2 would generate some kind of sinus-wave(ac) at the v0 node because of this "paradox". v0 would go from 2v to 0v to 2v and then back again, oscillating?
@@eyeballs97 assuming v+ was 1v and v- was 1v-6uV (there is a very very small difference between v+ and v-), then the amplified output would be 6v (assuming A=1,000,000). In the example, if Vo increased, it caused v- to increase, which will then cause Vo to decrease back to 6V (Vo's original value), this then makes v- decrease back to 1-6uV. At this point all voltages are back to its initial condition, so there wont be any oscillations. Only when Vo is once affected, then will it cause v- to change accordingly.
Great explanation. The concrete example is makes it easier to understand.
'What it does for us is it allows us to exploit the massive voltage gain to create extremely stable and nicely controlled sir cute' - fav part
THANK YOU. MY LECTURER COULD NEVER.
wonderful lesson very easy to understand and sink in the concepts! Well Done
How does v- go up when Vo went up?
by its definition : inverting input
So, since the difference on the inputs is amplified and by a huge number, do you need R1=R2 almost exactly for feedback to ensure that V+ = V- and you don't this cause saturation?
i don't get it. Why would u_o go up when the input voltage difference is 0. since the input and output voltage are linear with each other, u_o should become 0 and then you would have this oscillation of values back and forth. please explain
great explanation
thank you so much
best explanation !
What is the naem of this course please ?
in which amplifier you are feeding back a part of the processed signal to the input side
BRUH I OFTEN GET MY PROGRESS DESTROYED
why when v- goes up, v- at voltage divider mid point does not increase ?
So do inverting amps have negative feedback? When the inverting amp was drawn, there was no voltage divider drawn with it.
Thank you!!!!!!!!!
at 2:19 you said on the previous video ? what is the title of that video please
www.khanacademy.org/science/electrical-engineering/ee-amplifiers/ee-opamp/v/ee-feedback
Do note its developed 100 years ago !
macam mana nak buat?
So what happens after Vout decreases would it not just cause V- to decrease causing Vout to increase and then just be stuck in a loop of increasing and decreasing?
Vout will decrease only when the input voltage decreases, they are not inversely proportional to each other, they are just in opposite phase.
You have the idea right. Any change to Vout causes V- to change, which pushes Vout in the opposite direction of its change. That is the essence of feedback. In real life, Vout will settle at exactly the voltage you want and will stay there. (It may waver a tiny bit, but too small to notice.)
Sexcyvidoes
Thank you
Who can tell me what is the last video names and what is the next video name ,I just confused its order with this series
Why isn't the Non-Inverting Amplifier equation for the voltage is Vo/Vi = (1 + R2/R1) ?
The resistors have the same value, so (1 + R2/R1) = (1 + 1) = 2
Vo/Vi = 2 => Vo = 2*Vi
v+ =1 V
v- =1 V
Shouldn't the "Vo" be zero?
v+ and v- are APPROXIMATELY 0 due to the virtual ground theory but they are actually a very very small number... but when the difference between the small number is multiplied by A then we get Vout.
+Hillary Lastname 0967777302
Hillary Lastname but then that does not explain why the output stays at 2v
ATM Jahid Hasan it's not exactly, remember that the Vo=2Vm was obtained by approximating A as large (see the previous video) so that we ignored a small corrective term the actual v- differes from 1 by something of order 1/A which is very small.
No it wont be zero , this would be true for open loop configuration . As soon as we add feedback this formula Vout = gain*(V+ - V-) no longer applies
very helpful.
THANK YOU!!!
Cant thank you enough, seriously
if v- increases then v out decreases! isin't it?
Saptaki Debnath , yes that is besically making feedback.If you want to know more , you can check this video th-cam.com/video/nFD8O1Lp_1k/w-d-xo.html
BTA 41 feedback control diagram
thankyou sir
I am happy that I was able to know why we use negative feedback and how it helps to stabilize..
Thanks a lot.
do you reckon you could make a video on multiplexers?
Yes Sir Khan please bring forth deeper lessons for us please... Multiplexer would be nice.
Great tutorial we love it!
thank
Mind= blown 😳
this melinda crimeal life not me
Cindy Robinsin m
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Got a circuit you wanna build? Test it out here first. Browse for androidcircuitsolver on google
You are not Sal!!
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رحمان بکرا ٹی وی چینل منٹگمری۔۔۔۔Calll baka TV and
Bharat pagi
Hi
your explanation is not good enough to understand for starters
PEBCAC
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